Thread: Math Question of the Week

I just thought I would post some math problems for those who might enjoy them. Some may be very difficult, some may be very easy.Two hikers walking at the same rate leave a camp at the same time, one going due north, the other due east. After each has gone a mile, the northbound camper turns 15° clockwise and the eastbound hiker turns 15° counterclockwise. The repeat these turns each successive mile until they meet. How far from camp are they when they meet?

And we'll throw in a calc one too ...
The velocity (in feet/sec) of a a rocket whose initial mass (including fuel) is m is:

Code:

v = gt + u ln[ m / (m-rt)];  t < m / r

where u is the sxpulsion speed of the fuel, r is the rate at which the fuel is consumed, and g = -32 feet/sec² is the acceleration due to gravity. Find the position equation for a rocket wfor which m = 50,000 lbs, u = 12,000 feet/sec, and r = 400 lbs/sec. What is the height of the rocket when t = 100 sec? (Assuming the rocket was fired from ground level straight up)[

ok, this will just give me a headache

Code:

v = gt + u ln[ m / (m-rt)];  t < m / r

Thats supposed to mean something?

Originally Posted by sploit

where u is the sxpulsion speed of the fuel, r is the rate at which the fuel is consumed, and g = -32 feet/sec² is the acceleration due to gravity. Find the position equation for a rocket wfor which m = 50,000 lbs, u = 12,000 feet/sec, and r = 400 lbs/sec. What is the height of the rocket when t = 100 sec? (Assuming the rocket was fired from ground level straight up)

You know, I never even made it to geometry.

Someday someone is going to join this forum and figure this problem out. I'm pretty sure I could, just too lazy

It won't be me!

I had to figure both out. The first one was in trig, and the second, obviously was in calc.
I forget the second, although I could probably figure it out if i tried.
And for the second you pretty much just have to integrate the whole formula with respect to v, and then (v)(t) = height. So it's not that hard.

So it's not that hard.

'Are you sure?

Originally Posted by sploit

Code:

v = gt + u ln[ m / (m-rt)];  t < m / r

where u is the sxpulsion speed of the fuel, r is the rate at which the fuel is consumed, and g = -32 feet/sec² is the acceleration due to gravity. Find the position equation for a rocket wfor which m = 50,000 lbs, u = 12,000 feet/sec, and r = 400 lbs/sec. What is the height of the rocket when t = 100 sec? (Assuming the rocket was fired from ground level straight up)[

I'm not sure if I calculated it right. I got approxiamately 8800 feet.

Originally Posted by sploit

Two hikers walking at the same rate leave a camp at the same time, one going due north, the other due east. After each has gone a mile, the northbound camper turns 15° clockwise and the eastbound hiker turns 15° counterclockwise. The repeat these turns each successive mile until they meet. How far from camp are they when they meet?
[

If I did it right, they are 6 miles away. I may be off a tenth.

i'd say they never meet(which shows how sucky i am at trig)

It is approximately 6.08 miles from camp.
All you need to do is figure out that they will meet after 6 turns (totaling 90deg). Once you do that you can draw 6 triangles and using sin's and cos's and pythag thm. You can find the distance. I think we did this in geometry in 9th grade.

Originally Posted by sploit

It is approximately 6.08 miles from camp.
All you need to do is figure out that they will meet after 6 turns (totaling 90deg). Once you do that you can draw 6 triangles and using sin's and cos's and pythag thm. You can find the distance. I think we did this in geometry in 9th grade.

Hey, welcome back sploit

They way I figure it, after 6 turns, they'll both be heading East... they won't meet until they've made 12 turns.

The radius of the shape they form is 1/(2 tan7.5°), or 3.8 miles. They meet two radiuses (radii, if you prefer) away - 7.6 miles.

v = gt + u ln[ m / (m-rt)]

Integrating, I got:

<pre> g.t^2 / 2 - m^2.u / (r^2.t - m.r)</pre>...but I don't think that's right, because I'm getting funny results aftwerward.

I first changed:
<pre> ln[ m / (m-rt)]</pre>to:<pre> - ln(1 - rt/m)</pre>then used a substitution of z = 1-rt/m to integrate... have I made a mistake?

Originally Posted by Pete

They way I figure it, after 6 turns, they'll both be heading East... they won't meet until they've made 12 turns.

Hi Pete, welcome.

Both of them start off moving with 90 deg (due north & due east) difference & make 15 deg turns towards each other for each mile. 6 turns would be sufficient.

Originally Posted by sploit

I just thought I would post some math problems for those who might enjoy them. Some may be very difficult, some may be very easy.Two hikers walking at the same rate leave a camp at the same time, one going due north, the other due east. After each has gone a mile, the northbound camper turns 15° clockwise and the eastbound hiker turns 15° counterclockwise. The repeat these turns each successive mile until they meet. How far from camp are they when they meet?

And we'll throw in a calc one too ...
The velocity (in feet/sec) of a a rocket whose initial mass (including fuel) is m is:

Code:

v = gt + u ln[ m / (m-rt)];  t < m / r

where u is the sxpulsion speed of the fuel, r is the rate at which the fuel is consumed, and g = -32 feet/sec² is the acceleration due to gravity. Find the position equation for a rocket wfor which m = 50,000 lbs, u = 12,000 feet/sec, and r = 400 lbs/sec. What is the height of the rocket when t = 100 sec? (Assuming the rocket was fired from ground level straight up)[

I'll answer this one when you guys move into the 21st Century and use proper, easily interchangeable units. Feet per second!? Pounds weight!?!?

:P

Ah... For some reason I had them starting nouth and south. Must be that reading disorder...

Silas, can you integrate the function (in which units are irrelevant)? I'm interested in whether I'm right so far.

(Yes, I know it's stomach turning... particularly when pounds are coerced into a unit of mass).

heyhey.. to make that problem weirder you can imagine the camp was at the north or south pole then who would be moving?

Quote:
So it's not that hard.

'Are you sure? Laughing

I agree! Can't you post some "easy" ones? I haven't even had geometry.

Originally Posted by The_Science_Geek

Quote:
So it's not that hard.

'Are you sure? Laughing

I agree! Can't you post some "easy" ones? I haven't even had geometry.

whats 1+1?

Originally Posted by wallaby

Originally Posted by The_Science_Geek

Quote:
So it's not that hard.

'Are you sure? Laughing

I agree! Can't you post some "easy" ones? I haven't even had geometry.

whats 1+1?

11. Where's my Fields Medal?

The first one was fun. Haven’t tried the other one yet. Here is how I got the solution, to all of you who are stuck...
Well you can see it in the pic (at the bottom), but it may look confusing. In the origin of the graph (0,0) we have the camp. The red graph is the north-going man, and the blue is the east-going man. The green line is a 45° line, showing that the 2 men will meet after an exact integer of miles. Because I know this, I can keep the calculations to one of the men only - I've chosen the north going one.

I've used vector functions and/or summation, as you can see. The looong thing right under the graph describes the exact location of the north going man, after 7 miles. As you can see, the long expression is made up by 7 "blocks", each describing the position of the north going man after a full mile. As it turns out, after 7 miles, his north-position is equal to his east-position. That means that he now stands on the green 45° line. Because the route of the east-going man is a reflection of the route of the north going man in the 45° line, we know for sure that both men will be at that position after 7 miles. Use Pythagoras, and the distance from camp can be found (6.078 miles).

Another way to write it is using summation. Look top right, where I have two summations, which must be equal (the first one describes the north-position and the other the east-position of the north going man). As it turns out, m = 6 for the equation to fit. Or m = -1, but the man doesn't go backwards... To check we set m = 6 and once again we get the same 4.298 miles. Do the Pythagoras, and the distance is found once more.

I hope that helped someone

Originally Posted by sploit

I just thought I would post some math problems for those who might enjoy them. Some may be very difficult, some may be very easy.Two hikers walking at the same rate leave a camp at the same time, one going due north, the other due east. After each has gone a mile, the northbound camper turns 15° clockwise and the eastbound hiker turns 15° counterclockwise. The repeat these turns each successive mile until they meet. How far from camp are they when they meet?

Since they are walking at exactly the same speed, then they will meet after 12 turns. If they are not they may never meet.

Each walker is describing a path that is a regular 24-gon wih unt edges, but the easbound hiker followiing one with the initial edge along the x-axis and the northbound hiker one with the initial edge along the y axis. By symmetry thise 24-gons intersect at the origin and after 12 turns.

Originally Posted by sploit

And we'll throw in a calc one too ...
The velocity (in feet/sec) of a a rocket whose initial mass (including fuel) is m is:

Code:

v = gt + u ln[ m / (m-rt)];  t < m / r

where u is the sxpulsion speed of the fuel, r is the rate at which the fuel is consumed, and g = -32 feet/sec² is the acceleration due to gravity. Find the position equation for a rocket wfor which m = 50,000 lbs, u = 12,000 feet/sec, and r = 400 lbs/sec. What is the height of the rocket when t = 100 sec? (Assuming the rocket was fired from ground level straight up)[

Integrate the rocket equaton to find that the height h is given by

h =

You can plut in the given values if you wish to find the specific height at 100 sec.